# Thread: explain why these two expressions are equal

1. ## explain why these two expressions are equal

sin(9pi/7) = -sin(2pi/7)

Stuck....

Step by step would be appreciated very much.

Thanks

2. ## Re: explain why these two expressions are equal

note ...

$\sin\left(\frac{9\pi}{7}\right) = \sin\left(\pi + \frac{2\pi}{7}\right)$

... now use the sum identity for sine

3. ## Re: explain why these two expressions are equal

If you draw the terminal angles $\frac{9\pi}{7}$ and $\frac{2\pi}{7}$ on a unit circle, you will see that they differ by $\pi$. The sine of the angle is the y-coordinate of that angle/point on the unit circle. You will see that one is the negative of the other.

In general, $\sin (x + \pi) = -\sin x$.

4. ## Re: explain why these two expressions are equal

Originally Posted by skeeter
note ...

$\sin\left(\frac{9\pi}{7}\right) = \sin\left(\pi + \frac{2\pi}{7}\right)$

... now use the sum identity for sine
Why use the sum identity? You already have it in the form \displaystyle \begin{align*} \sin{\left(\pi + \theta \right)} \equiv -\sin{(\theta)} \end{align*}.

5. ## Re: explain why these two expressions are equal

Originally Posted by Prove It
Why use the sum identity? You already have it in the form \displaystyle \begin{align*} \sin{\left(\pi + \theta \right)} \equiv -\sin{(\theta)} \end{align*}.
so, you assume the OP is familiar with that specific identity derived from the sum identity?

6. ## Re: explain why these two expressions are equal

Originally Posted by skeeter
so, you assume the OP is familiar with that specific identity derived from the sum identity?
While that identity can be derived from the sum identity, it is also blatantly obvious from the unit circle.

7. ## Re: explain why these two expressions are equal

Originally Posted by Prove It
While that identity can be derived from the sum identity, it is also blatantly obvious from the unit circle.
"blatantly obvious from the unit circle" ... thanks for the lesson.

8. ## Re: explain why these two expressions are equal

i am appreciate you ... this is right .... work smart not hard...

\sin\left(\frac{9\pi}{7}\right) = \sin\left(\pi + \frac{2\pi}{7}\right)